A square steel plate is 9.00 cm on a side and 0.600 cm thick. Assume that the plate is oriented with the square faces horizontal to the ground.
A. Find the shear strain that results if a force of magnitude 1.00×106 N is applied to the top square face of the plate, parallel to the side, and a force of equal magnitude is applied in the opposite direction to the bottom face of the plate as shown .
Note that the shear modulus of steel in pascals is 7.50×1010 Pa.
B. Find the displacement Delta x in centimeters.
help on this plz
To find the shear strain and displacement in this scenario, we can apply the appropriate formulas and calculations.
A. Shear Strain Calculation:
Shear strain (ɛ) is given by the formula:
ɛ = (
Where:
F is the magnitude of the force applied (1.00x10^6 N in this case),
A is the cross-sectional area of the plate,
G is the shear modulus of steel.
To find the cross-sectional area of the plate:
The plate is square with sides measuring 9.00 cm. Since the plate is oriented with the square faces horizontal to the ground, the cross-sectional area is given by the formula:
A = L^2
Where:
L is the length of each side of the square plate (9.00 cm in this case).
So, we have:
A = (9.00 cm)^2 = 81.00 cm^2
Now we can substitute the values into the shear strain formula:
ɛ = (1.00x10^6 N / 81.00 cm^2) / (7.50x10^10 Pa)
Simplifying this expression gives us the shear strain.
B. Displacement Calculation:
The displacement (Δx) is given by the formula:
Δx = (F / A) / (2G)
Using the same values for F, A, and G as in the shear strain calculation, we can substitute them into the displacement formula:
Δx = (1.00x10^6 N / 81.00 cm^2) / (2 * 7.50x10^10 Pa)
Simplifying this expression gives us the displacement Δx in centimeters.